Question

In the xy coordinate plane the graph of the equations y = 3x^2 -12x -36 has zeros at x __ and x = ___
The graph has a minimum at (. ___, -48)
What are the values of x

Answer 1

To find the zeros of the equation \( y = 3x^2 - 12x - 36 \), we can use the quadratic formula or factor the equation.

### Step 1: Find the zeros

The standard form of a quadratic equation is:

\[
ax^2 + bx + c = 0
\]

In this case, we have:

- \( a = 3 \)
- \( b = -12 \)
- \( c = -36 \)

To find the zeros, we set the equation to zero:

\[
3x^2 - 12x - 36 = 0
\]

We can divide the entire equation by 3 to simplify:

\[
x^2 - 4x - 12 = 0
\]

### Step 2: Factor the quadratic

Next, we need to factor \( x^2 - 4x - 12 \). We look for two numbers that multiply to \(-12\) (the constant term) and add to \(-4\) (the coefficient of \(x\)).

The correct factors are \( -6 \) and \( 2 \):

\[
(x - 6)(x + 2) = 0
\]

### Step 3: Solve for x

Setting each factor equal to zero gives us the zeros:

1. \( x - 6 = 0 \) ⇒ \( x = 6 \)
2. \( x + 2 = 0 \) ⇒ \( x = -2 \)

Thus, the zeros of the equation are \( x = -2 \) and \( x = 6 \).

### Step 4: Find the minimum point

The vertex of a quadratic function \( y = ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \).

For our equation:

\[
x = -\frac{-12}{2 \cdot 3} = \frac{12}{6} = 2
\]

Now, we substitute \( x = 2 \) back into the original equation to find the corresponding \( y \)-value:

\[
y = 3(2)^2 - 12(2) - 36
\]
\[
= 3(4) - 24 - 36
\]
\[
= 12 - 24 - 36
\]
\[
= 12 - 60 = -48
\]

### Summary

The zeros are:
- \( x = -2 \)
- \( x = 6 \)

The vertex (minimum point) is:
- \( (2, -48) \)

Thus:

- The values of \( x \) (the zeros) are \( x = -2 \) and \( x = 6 \).
- The graph has a minimum at \( (2, -48) \).